Bài 6 trang 133 sgk đại số 11


Nội dung bài giảng

Bài 6. Tính:

\(\eqalign{
& a)\mathop {\lim }\limits_{x \to + \infty } ({x^4} - {x^2} + x - 1) \cr
& b)\mathop {\lim }\limits_{x \to - \infty } ( - 2{x^3} + 3{x^2} - 5) \cr
& c)\mathop {\lim }\limits_{x \to - \infty } (\sqrt {{x^2} - 2x + 5}) \cr
& d)\mathop {\lim }\limits_{x \to + \infty } {{\sqrt {{x^2} + 1} + x} \over {5 - 2x}} \cr} \)

Giải:

\(\eqalign{
& a)\mathop {\lim }\limits_{x \to + \infty } ({x^4} - {x^2} + x - 1) = \mathop {\lim }\limits_{x \to + \infty } {x^4}\left( {1 - {1 \over {{x^2}}} + {1 \over {{x^3}}} - {1 \over {{x^4}}}} \right) = + \infty \cr
& b)\mathop {\lim }\limits_{x \to - \infty } ( - 2{x^3} + 3{x^2} - 5) = \mathop {\lim }\limits_{x \to - \infty } {x^3}\left( { - 2 + {1 \over x} - {5 \over {{x^2}}}} \right) = + \infty \cr
& c)\mathop {\lim }\limits_{x \to - \infty } (\sqrt {{x^2} - 2x + 5} ) = \mathop {\lim }\limits_{x \to - \infty } |x|\sqrt {1 - {2 \over x} + {5 \over {{x^2}}}} = + \infty \cr
& d)\mathop {\lim }\limits_{x \to + \infty } {{\sqrt {{x^2} + 1} + x} \over {5 - 2x}} = \mathop {\lim }\limits_{x \to + \infty } {{x\left( {\sqrt {1 + {1 \over {{x^2}}}} + 1} \right)} \over {5 - 2x}} = \mathop {\lim }\limits_{x \to + \infty } {{\left( {\sqrt {1 + {1 \over {{x^2}}}} + 1} \right)} \over {{5 \over x} - 2}} = - 1 \cr} \)